Multi-year analysis of solvency capital in life insurance

Abstract

With the commencement of the Solvency II directive, insurers in the European Union need to provide a projection of their solvency figures into the future (as part of the Own Risk and Solvency Assessment, ORSA). This is a highly complex task since future solvency figures depend on the development of numerous (stochastic) risk factors. The required evaluations are numerically challenging, which in practice forces companies to limit their analyses to only a few selected deterministic scenarios. These deterministic scenarios clearly cannot describe the full probability distribution of a company’s future solvency situation. The focus of this paper is on financial guarantees in participating life insurance products. In particular, we study two major types of interest rate guarantees in life insurance, a maturity guarantee and a (path-dependent) cliquet-style guarantee. In order to derive entire probability distributions of future solvency ratios, we limit the model framework to two sources of risk (a Hull–White model for interest rates and a geometric Brownian motion for stocks). This partly leads to closed-form solutions of the market-consistent valuation of the liabilities, ensures higher accuracy in computations and less numerical effort. Furthermore, the model allows for a detailed analysis of the impact of the different types of interest rate guarantees on the future solvency situation. Our results suggest that the type of guarantee has a significant impact on the long-term solvency of the company.

Appendix

We recall formula (5) from Sect. 5.4.

$$\eqalign{ \frac{{SC{R_t}}}{{{L_t}}} = & \frac{{{q_{99.5\% }}\left( {{E_t} - {e^{ - \int\limits_t^{t + 1} {r\left( s \right)ds} }}{E_{t + 1}}} \right)}}{{{L_t}}} = \frac{{{E_t}}}{{{L_t}}} - \frac{{{q_{0.5\% }}\left( {{e^{ - \int\limits_t^{t + 1} {r\left( s \right)ds} }}{E_{t + 1}}} \right)}}{{{L_t}}} \cr = & \frac{{{A_t}}}{{{L_t}}} - 1 - {q_{0.5\% }}\left( {{e^{ - \int\limits_t^{t + 1} {r\left( s \right)ds} }}\frac{{{A_{t + 1}}}}{{{L_t}}} - {e^{ - \int\limits_t^{t + 1}r\left( s \right)ds}}\frac{{{L_{t + 1}}}}{{{L_t}}}} \right) \cr}$$

It remains to show that the first part of the argument in the quantile can be written as a product of ${A}_{t}/{L}_{t}$ and some stochastic factor, namely,

$$\begin{aligned}&{e}^{-\int_{t}^{t+1}r\left(s\right)ds} \frac{{A}_{t+1}}{{L}_{t}}=\frac{{A}_{t}}{{L}_{t}}\mathrm{exp}\left\{{x}_{S}{\lambda }_{S}-\frac{1}{2}{\left({x}_{S}{\sigma }_{S}\right)}^{2}+\left({x}_{S}{x}_{B}{\sigma }_{S}{\sigma }_{r}\rho -{x}_{B}{\lambda }_{r}\right)\right. \\ & \left.\quad\int_{t}^{t+1}\left({\sum }_{j=1}^{{T}^{*}}{x}_{tj}B\left(s,t+j\right)\right)ds-\frac{1}{2}{\left({x}_{B}{\sigma }_{r}\right)}^{2}\int_{t}^{t+1}{\left({\sum }_{j=1}^{{T}^{*}}{x}_{tj}B\left(s,t+j\right)\right)}^{2}ds\right. \\ & \left.\quad +{x}_{S}{\sigma }_{S}\left[\rho \left({\tilde{W}}_{1}\left(t+1\right)-{\tilde{W}}_{1}\left(t\right)\right)+\sqrt{1-{\rho }^{2}}\left({\tilde{W}}_{2}\left(t+1\right)-{\tilde{W}}_{2}\left(t\right)\right)\right]\right. \\ & \left.\quad-{x}_{B}{\sigma }_{r}\int_{t}^{t+1}\left({\sum }_{j=1}^{{T}^{*}}{x}_{tj}B\left(s,t+j\right)\right)d{\tilde{W}}_{1}(s)\right\}\end{aligned}$$

under the real-world measure $\mathcal{P}$. In the case of the maturity company, the second part corresponds to

$$\begin{aligned}&{e}^{-\int_{t}^{t+1}r\left(s\right)ds}\frac{ {L}_{t+1}}{{L}_{t}}={e}^{-\int_{t}^{t+1}r\left(s\right)ds}\frac{ P(t+1,T)}{{L}_{t}}{L}_{T}^{G}+{e}^{-\int_{t}^{t+1}r\left(s\right)ds}\delta \alpha \\ &\quad\left[\frac{{A}_{t+1}}{{L}_{t}} N\left({d}_{1}(t+1)\right)-\frac{{L}_{T}^{G}}{\alpha } \frac{P\left(t+1,T\right)}{{L}_{t}} N\left({d}_{2}(t+1)\right)\right] ,\end{aligned}$$

where ${e}^{-\int_{t}^{t+1}r\left(s\right)ds} {A}_{t+1}/{L}_{t}$ was already discussed above and

$${e}^{-\int_{t}^{t+1}r\left(s\right)ds}\frac{ P(t+1,T)}{{L}_{t}}={I}_{P}\cdot {II}_{P}\cdot \frac{P\left(t,T\right)}{{L}_{t}}={I}_{P}\cdot {II}_{P}\cdot \frac{P\left(t,T\right)}{{A}_{t}}\frac{{A}_{t}}{{L}_{t}} ,$$

$$\begin{aligned}{I}_{P}=\,& \mathrm{exp}\left\{\frac{1}{a}\left(\alpha \left(t\right)+\frac{{\lambda }_{r}}{a}\right)\left(1-{e}^{-a}\right)-\frac{{\lambda }_{r}}{a}-\mathrm{ln}\left(\frac{{P}^{M}\left(0,t\right)}{{P}^{M}\left(0,t+1\right)}\right)\right. \\ & \left.-\frac{{\sigma }_{r}^{2}}{2{a}^{2}}\left(1+\frac{2}{a}{e}^{-a\left(t+1\right)}\left(1-{e}^{a}\right)-\frac{1}{2a}{e}^{-2a\left(t+1\right)}\left(1-{e}^{2a}\right)\right)\right. \\ & \left.+B\left(t+1,T\right)\left({e}^{-a}\alpha \left(t\right)-\alpha \left(t+1\right)-\frac{{\lambda }_{r}}{a}\left(1-{e}^{-a}\right)\right)\right\}\cdot \frac{A\left(t+1,T\right)}{A\left(t,T\right)} ,\\ I{I}_{P}= \,& \mathrm{exp}\left\{-\frac{{\sigma }_{r}}{a}\left({\tilde{W}}_{1}\left(t+1\right)-{\tilde{W}}_{1}\left(t\right)\right)+\int_{t}^{t+1}{e}^{a\left(u-\left(t+1\right)\right)}d{\tilde{W}}_{1}\left(u\right)\left(\frac{{\sigma }_{r}}{a}-B\left(t+1,T\right){\sigma }_{r}\right)\right\} .\end{aligned}$$

${I}_{P}$ is deterministic and $I{I}_{P}$ is stochastic. ${d}_{1}(t+1)$ and ${d}_{2}(t+1)$ entail ${A}_{t+1}/P(t+1,T)$, which can be rewritten as

$$\frac{{A}_{t+1}}{P(t+1,T)}=\frac{ {e}^{-\int_{t}^{t+1}r\left(s\right)ds} {A}_{t+1}/{L}_{t}}{{e}^{-\int_{t}^{t+1}r\left(s\right)ds}P(t+1,T)/{L}_{t}}$$

and was therefore already covered above. It now remains to show that $P(t,T)/{A}_{t}$ is fully determined by ${A}_{t}/{L}_{t}$. The relation between the two is given by

$$\frac{{L}_{t}}{{A}_{t}}=\frac{P(t,T)}{{A}_{t}}{L}_{T}^{G}+\delta \alpha N\left({d}_{1}\left(t,\frac{P\left(t,T\right)}{{A}_{t}}\right)\right)-\delta {L}_{T}^{G}\frac{P\left(t,T\right)}{{A}_{t}}N\left({d}_{2}\left(t,\frac{P\left(t,T\right)}{{A}_{t}}\right)\right) .$$

We are thus interested in the existence of an inverse function of

$$\frac{{L}_{t}}{{A}_{t}}\left(x\right)=x{L}_{T}^{G}+\delta \alpha N\left({d}_{1}\left(t,x\right)\right)-\delta {L}_{T}^{G}xN\left({d}_{2}\left(t,x\right)\right) ,$$

$$\eqalign{ & {d_1}\left( {t,x} \right) = \frac{{{\text{ln}}\left( {\frac{1}{{Kx}}} \right) + \frac{1}{2}\left( {{\Sigma ^2}\left( T \right) - {\Sigma ^2}(t)} \right)}}{{\sqrt {{\Sigma ^2}\left( T \right) - {\Sigma ^2}(t)} }}, \cr & {d_2}\left( {t,x} \right) = {d_1}\left( {t,x} \right) - \sqrt {{\Sigma ^2}\left( T \right) - {\Sigma ^2}\left( t \right)} . \cr}$$

${L}_{t}/{A}_{t}(x)$ is defined for all positive $x$. Clearly, the function is continuous on its domain as all of its components are continuous. Next, we check for monotonicity by taking the derivative.

$$\begin{aligned}\frac{\partial \frac{{L}_{t}}{{A}_{t}}(x)}{\partial x}=\, &{L}_{T}^{G}-\delta \alpha {N}^{{\prime}}\left({d}_{1}\left(t,x\right)\right)\frac{1}{\sqrt{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}}\frac{1}{x}\\ &-\delta {L}_{T}^{G}N\left({d}_{2}\left(t,x\right)\right)+\delta {L}_{T}^{G}{N}^{{\prime}}\left({d}_{2}\left(t,x\right)\right)\frac{1}{\sqrt{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}}\end{aligned}$$

(6)

since

$$\frac{\partial {d}_{1}(t,x)}{\partial x}=\frac{\partial {d}_{2}(t,x)}{\partial x}=-\frac{1}{\sqrt{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}}\frac{1}{x} .$$

$N{^{\prime}}(x)$ is the probability density function of a standard normal random variable, that is

$${N}^{{\prime}}\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2} .$$

The evaluation of this function at ${d}_{1}(t,x)$ and ${d}_{2}(t,x)$ yields

$$\begin{aligned}{N}^{{\prime}}\left({d}_{1}\left(t,x\right)\right)&=\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{1}{2}\frac{1}{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}{\left[\mathrm{ln}\left(\frac{1}{Kx}\right)+\frac{1}{2}\left({\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)\right)\right]}^{2}\right\} \\ &=\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{1}{2}\frac{1}{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}{\left(\mathrm{ln}\left(\frac{1}{Kx}\right)\right)}^{2}-\frac{1}{2}\mathrm{ln}\left(\frac{1}{Kx}\right)-\frac{1}{8}\left({\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)\right)\right\} ,\\ {N}^{{\prime}}\left({d}_{2}\left(t,x\right)\right)&=\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{1}{2}\frac{1}{{\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)}{\left[\mathrm{ln}\left(\frac{1}{Kx}\right)-\frac{1}{2}\left({\Sigma }^{2}\left(T\right)-{\Sigma }^{2}\left(t\right)\right)\right]}^{2}\right\} \\ &={N}^{{\prime}}\left({d}_{1}\left(t,x\right)\right)\frac{1}{Kx} .\end{aligned}$$

Since $K$ is the strike price ${L}_{T}^{G}/\alpha$, the two parts of (6) containing $N{^{\prime}}$ cancel out. We finally obtain

$$\frac{\partial \frac{{L}_{t}}{{A}_{t}}(x)}{\partial x}={L}_{T}^{G}-\delta {L}_{T}^{G}N\left({d}_{2}\left(t,x\right)\right)={L}_{T}^{G}\left(1-\delta N\left({d}_{2}\left(t,x\right)\right)\right)>0$$

as long as $\delta$ does not exceed one. In total, the function ${L}_{t}/{A}_{t}(x)$ is continuous and strictly monotonically increasing on its domain of all positive $x$. Checking the limits, it takes values in $\left(\delta \alpha ,+\infty \right)$. Therefore, an inverse function exists on the domain $\left(\delta \alpha ,+\infty \right)$, which is equivalent to $P(t,T)/{A}_{t}$ being uniquely determined by ${A}_{t}/{L}_{t}$.